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I have an impulse response of an LTI system which is

h(n)= n*u(n)-u(n-2) where n=[0,3] (1)

Now regarding the stability and causality of this system, i've drawn the conclusion that the system is causal since the output only relies on past or present values of n and it is stable since n is bounded. In this case n=0,1,2,3. Correct me here if i'm wrong but in case h(n) was u(n) - u(n-2) without knowing anything about n, it would stil be stable right? And if h(n) is the same as the original (1) but again without knowing anything about n it would be unstable correct?

Now i want to analyze this h(n) in odd and even components and also find the linear difference equation of h(n) ,but i can't find anything in my textbook that is of any help. It's the first time i've attended a signals and systems course and i'm really struggling to understand a few things, so any help here would be greatly appreciated.

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  • $\begingroup$ Is $h[n]$ given by $n\cdot (u[n]-u[n-2])$ or by $n\cdot u[n] - u[n-2]$? $\endgroup$
    – Matt L.
    Jul 7, 2020 at 16:06
  • $\begingroup$ @MattL. second one $\endgroup$
    – Notoriousphd
    Jul 7, 2020 at 16:22

1 Answer 1

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You're right about your conclusions. And now a few hints to help you solve the questions:

  1. Your textbook should state somewhere the definition of even and odd parts of a (real-valued) sequence: $$h_e[n]=\frac12\big(h[n]+h[-n]\big)\\h_o[n]=\frac12\big(h[n]-h[-n]\big)$$

  2. In this case (FIR filter), the difference equation is simply given by the convolution of the input sequence $x[n]$ with the impulse response $h[n]$: $$y[n]=\sum_kx[k]h[n-k]$$

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  • $\begingroup$ 1) Yes you are right. This was so early in the book(like page 9) i don't know how i missed it.Regarding (2) how do i know this is a FIR filter not a IIR one? Is it because i have discrete system, or because it's stable(meaning n is bounded)? $\endgroup$
    – Notoriousphd
    Jul 7, 2020 at 22:39
  • $\begingroup$ FIR stands for Finite impulse response. Since h(n) is not infinite it is FIR. It is only between 0 and 3 $\endgroup$
    – VMMF
    Jul 8, 2020 at 0:00

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